Question

prove that angles opposite to equal sides of an isosceles triangle are equal​

Answer 1

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Answer 2

Answer:

Hey mate, this is your answer.

Step-by-step explanation:

Take a triangle ABC, in which AB = BC

Construct AC bisector of A meeting BC at P.

In ∆ABP and ∆ACP

AP = AC [common]

AB = AC [given]

Angle BAP = angle CAP [by construction]

Therefore, ∆ABP concurrent ∆ACP [S. A.S. ]

This implies, angle ABP = angle ACP [C. P. C. T. ]

Hence proved that angles opposite to equal sides of a triangle are equal.

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